I Is mass the source of spacetime?

[Angelika10]

In this article:
https://link.springer.com/article/10.1140/epjc/s10052-017-4695-y
they explain that introducing new fields to the EFE and introducing unknown particles is equivalent. Therefore, modifying GR is not an alternative to dark matter, its conceptually the same.

 

[PeterDonis]

Newton perfectly fits for the solar system, but not for whole galaxies.

This is not correct. Newtonian gravity with an appropriate mass distribution can fit galaxy rotation curves just fine. Newtonian gravity is not limited to point masses with an inverse square gravitational force.

What is the difference between the solar system and the whole galaxy?

The mass distribution; the solar system is well approximated by a small number of point masses interacting with an inverse square force. The galaxy as a whole is not; it is better approximated by a continuous mass distribution with a density that varies with position. (Strictly speaking, one could try to model a galaxy as a large number of interacting point masses, but this is computationally intractable because the number of point masses required would be so large--a hundred billion or more.)

The galaxy is not moving so fast around a bigger mass.

Yes, it is. Our galaxy is part of the Local Group, which in turn is part of a larger galaxy cluster, which in turn is part of a supercluster.

What follows from the movement of the solar system around the galaxy center?

Um, the fact that the galaxy is composed of a large gravitating mass?

I'm not sure what you're trying to say here.

 

[PeterDonis]

they explain that introducing new fields to the EFE and introducing unknown particles is equivalent.

That is basically correct, but it doesn't mean what you think it means. See below.

Therefore, modifying GR is not an alternative to dark matter, its conceptually the same.

Wrong. Modifying GR means modifying the EFE itself. Dark matter just means modifying the stress-energy distribution, not the EFE.

 

[vanhees71]

Yes, the EM field can be interpreted as a curvature in an internal, abstract "space". But that is not the same as saying that the EM field is curved. It's the internal abstract space that is curved (in this interpretation).

Yes, but mathematically it's the same concept, namely gauge invariance. GR can also be understood as a gauge theory. The global symmetry gauged is the Poincare group, i.e., the symmetry group of Minkowski space, and that's why in this case the connection, pseudo-metric, curvature etc. refer to the spacetime geometry and not to an "internal abstract space" (a fiber bundle).

 

[rc1]

And the Galaxy center? We're moving with 220km/s around the galactic center, in comparison to 30 km/s Earth around the sun.
Therefore, the field of the galactic center is big at our place!
But, since we're moving in free fall (with the solar system), I would suppose we do not measure any derivation from the flat space in the solar system.

Angelika10 said- "Therefore, the field of the galactic center is big at our place!"- Newton's 1st Law (of linear motion)- Speed is not dependent on force. In this case the motion is rotational motion w= v/r. Centripetal Force = mv^2/r and this is proportional (equal) to the field strength. Though the masses are the same the radii are different. The Earth is 150 M km from the Sun but much further from the Galactic Core (27,000 ly- 1 ly = 10 T km= 10x 10^12 km). So the greater the radius the smaller the Centripetal Force.
In fact if the field due to the Galactic core was greater than the Sun I suspect that the Galactic tidal forces would tear apart the Solar System.

I could be wrong... this is just a classical analysis.

 

[rc1]

You could probably argue that spacetime is approximately flat on the scale of the Solar System.

 

[Ibix]

You could probably argue that spacetime is approximately flat on the scale of the Solar System.

No you couldn't - the planets wouldn't orbit. You could argue that the contribution to the total curvature from masses outside the system is negligible on the solar system spatial scale and thousand year timescale.

 

[rc1]

No you couldn't - the planets wouldn't orbit. You could argue that the contribution to the total curvature from masses outside the system is negligible on the solar system spatial scale and thousand year timescale.

Yes sorry that's what I meant. Thanks Ibix for the correction.

Also Murray Gell- Mann has a TED Talk on why physics is beautiful at different scales because it's similar.

 

[Nugatory]

In fact if the field due to the Galactic core was greater than the Sun I suspect that the Galactic tidal forces would tear apart the Solar System.

Don’t think so... if two gravitating bodies produce the same field strength at a given point, the tidal effects are weaker for the larger and more distant one.

(No GR needed for this calculation, the Newtonian approximation is fine).

 

[Angelika10]

The Ricci tensor is not on the LHS of the field equations; the Einstein tensor is. The Einstein tensor is not exactly "volume gain"; but even leaving that aside, what the LHS of the EFE represents is not just "volume gain" but "volume gain or loss". For normal matter with "attractive gravity", the EFE predicts volume loss, not gain. (More precisely, the volume of a small ball of test particles will decrease.)

This paper by Baez and Bunn gives a good basic treatment:

https://arxiv.org/abs/gr-qc/0103044

ok, they write, that a volume of a small ball of testparticles will decrease in time. And that's the basic meaning of "gravity attracts".

My question is another: If I the mass is increased, will the volume increase or decrease because of the additional mass?
... can the metric tensor help? diag(B, -A, -r², -r²sin(theta))

 

[PeterDonis]

they write, that a volume of a small ball of testparticles will decrease in time. And that's the basic meaning of "gravity attracts".

More precisely, that the normalized "acceleration" of the volume--the second derivative of the volume with respect to time, divided by the volume--is negative. That means the volume, if it starts from "rest" (zero rate of change with time), will decrease at a rate that increases with time.

If I the mass is increased, will the volume increase or decrease because of the additional mass?

The Ricci tensor does not describe "mass". It describes density of stress-energy. For the simple case of a perfect fluid, the Ricci tensor--more precisely, the piece of it that determines the (negative) "acceleration" of the volume described above--will be $\rho + 3 p$, where $\rho$ is the energy density of the fluid and $p$ is the pressure. So positive #\rho + 3p## means the "acceleration" of the volume is negative--"attractive" gravity. Note that this is for an observer who is inside the fluid, observing the behavior of a small ball of test particles that is also inside the fluid.

can the metric tensor help? diag(B, -A, -r², -r²sin(theta))

Where is this metric tensor coming from?

 

[Angelika10]

The Ricci tensor does not describe "mass". It describes density of stress-energy. For the simple case of a perfect fluid, the Ricci tensor--more precisely, the piece of it that determines the (negative) "acceleration" of the volume described above--will be $\rho + 3 p$, where $\rho$ is the energy density of the fluid and $p$ is the pressure. So positive #\rho + 3p## means the "acceleration" of the volume is negative--"attractive" gravity. Note that this is for an observer who is inside the fluid, observing the behavior of a small ball of test particles that is also inside the fluid.

Hm, ok. I understand this. But the question is: if the energy-momentum-tensor is increased, what happens with the volume?

Where is this metric tensor coming from?

From the standard form of a static, spherically symetric metric.

$ds^2 = B(r)c^2dt^2 -A(r)dr^2 - r^2(d\theta^2+sin^2 \theta d\phi^2)$

so

$(g_{\mu\nu}) = diag (B(r), -A(r), -r^2, -r^2 sin^2\theta)$

 

[PeterDonis]

if the energy-momentum-tensor is increased

What does it even mean to "increase" a tensor? A tensor is not a number.

From the standard form of a static, spherically symetric metric.

A non-vacuum spacetime (i.e., a spacetime with a nonzero Ricci tensor, which requires a nonzero stress-energy tensor) will not necessarily be either static or spherically symmetric.

 

[vanhees71]

More precisely, that the normalized "acceleration" of the volume--the second derivative of the volume with respect to time, divided by the volume--is negative. That means the volume, if it starts from "rest" (zero rate of change with time), will decrease at a rate that increases with time.The Ricci tensor does not describe "mass". It describes density of stress-energy. For the simple case of a perfect fluid, the Ricci tensor--more precisely, the piece of it that determines the (negative) "acceleration" of the volume described above--will be $\rho + 3 p$, where $\rho$ is the energy density of the fluid and $p$ is the pressure. So positive #\rho + 3p## means the "acceleration" of the volume is negative--"attractive" gravity. Note that this is for an observer who is inside the fluid, observing the behavior of a small ball of test particles that is also inside the fluid.

But this is not the Ricci but the energy-momentum tensor, $T_{\mu \nu}=(\epsilon+P) u^{\mu} u^{\nu}-P g^{\mu \nu}$ (for an ideal fluid). The source of the gravitational field (or equivalently space-time curvature) is the energy-momentum-stress of the "matter", as described by the Einstein field equations,
$$R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu} = \kappa T_{\mu \nu}.$$
Here $R_{\mu \nu}$ is the Ricci tensor, $R$ its trace, and $\kappa=8 \pi G/c^4$ with $G$ being Newton's gravitational constant. The sign on the right-hand side depends on the convention, i.e., how you contract the Riemann curvature tensor to the Ricci tensor.

An equivalent form is
$$R_{\mu \nu}=\kappa \left (T_{\mu \nu}-\frac{1}{2} T g_{\mu \nu} \right), \quad T=T_{\mu}^{\mu}.$$

 

[PeterDonis]

this is not the Ricci but the energy-momentum tensor

In the paper by Baez and Bunn that I linked to, they rearrange the field equation so that the Ricci tensor is on the LHS, in the "equivalent form" you give at the end of your post. That is because the Ricci tensor is the tensor that has the direct physical interpretation being discussed (the "acceleration" of the volume, divided by the volume, of a small ball of test particles, caused by "mass" or more precisely density of stress-energy). That is what I was referring to.

 

[catlove]

Tenors (field vectors) are associated with an entity that consitutes matter yet in vacuum there is not matter so tenor theory (guess) is speculative.

 

[PeterDonis]

Tenors (field vectors) are associated with an entity that consitutes matter

Not necessarily. Spacetime itself is described by tensors (metric tensor, Riemann tensor, etc.) even in the absence of matter.

so tenor theory (guess) is speculative.

Your guess is incorrect.

 

[catlove]

Really?

"Maxwell's electrodynamics proceeds in the same unusual way already analyzed in studying his electrostatics. Under the influence of hypotheses which remain vague and undefined in his mind, Maxwell sketches a theory which he never completes, he does not even bother to remove contradictions from it; then he starts changing this theory, he imposes on it essential modifications which he does not notify to his reader; the latter tries in vain to fix the fugitive and intangible thought of the author; just when he thinks he has got it, even the parts of the doctrine dealing with the best studied phenomena are seen to vanish. And yet this strange and disconcerting method led Maxwell to the electromagnetic theory of light!" (Duhem, 1902).

 

[PeterDonis]

Really?

Yes.

Duhem, 1902

A reference from 119 years ago, particularly from a philosopher and not a scientist, more particularly one that makes incorrect claims (Maxwell electrodynamics is perfectly consistent), and even more particularly when the issue it is talking about has nothing whatever to do with tensors, is not a good basis for discussion.